Please help me. I have no idea what I'm doing.

The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i

ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

8 0

3 years ago

Is it a problem that the stress intensity factor, K, from Irwin's near-tip approximation approaches infinity as you get close to

valina [46]

Answer:

No it is not a problem

Explanation:

It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .

and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "

3 0

3 years ago

In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh

astraxan [27]

( 12 17 18 19 25 )

Explanation:

First Pass:

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

Second Pass:

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

Third Pass:

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

Fourth Pass:

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

Fifth Pass:

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0

3 years ago

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length

EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2 $l$ . The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2 $l$

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

so

D₃ = D $($ ΔP₁₋₂ / ΔP₂₋₃ $)^{\frac{1}{4}$

we substitute

D₃ = D $($ 1.657 $)^{\frac{1}{4}$

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0

3 years ago

The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1

Thepotemich [5.8K]

Answer:

a) The flow capacity of the fan is 3150 L/min

d) the minimum diameter is 0.11 m

Explanation:

given data:

A = area of residence = 200 m²

h = height of building = 2.7 m

Percentage of air that must be replaced by fresh air is 35%

v = velocity of air in the duct = 5.5 m/s

a) The volume of the entire building is:

$Volume=2.7*200=540m^{3}$

The flow capacity of the fan is equal to:

$Flow=\frac{0.35*540}{60} =3.15m^{3} /min$

$Flow=3.15\frac{m^{3} }{min} *\frac{1000L}{1m^{3} } =3150L/min$

b) The volume flow rate of fresh air is equal to:

$Flow=\frac{\pi *d^{2} }{4} V\\d=\sqrt{\frac{4*Flow}{\pi V} } =\sqrt{\frac{4*3.15}{\pi *5.5*60} } =0.11m$

6 0

4 years ago

Please help me. I have no idea what I'm doing.​

Please help me. I have no idea what I'm doing.